# The Hilbert space of non-relativistic QM

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What exactly is the Hilbert space of a massive spin 0 particle in non-relativistic QM? The following construction defines a Hilbert space H, but is it the right one? We could e.g use some subspace of H instead. And what if we use Riemann integrals instead of Lebesgue integrals?

Let G be the set of measurable functions $f:\mathbb R^3\rightarrow\mathbb C$ such that $|f|^2$ is Lebesgue integrable. Define an equivalence relation on G by saying that f and g are equivalent if the set on which f-g is non-zero has Lebesgue measure 0, and define H to be the set of such equivalence classes. Define the inner product of two arbitrary equivalence classes $\bar f$ and $\bar g$ by

$$\langle \bar f|\bar g\rangle=\int\limits_{-\infty}^\infty d^3x f^*(x) g(x)$$

where f and g are arbitrary functions in the equivalence classes $\bar f$ and $\bar g$.

dextercioby
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Your construction would lead to the L^2 (R^3) which can be shown to be a Hillbert space, actually the single (up to a unitary equivalence) infinite-dimensional separable Hilbert space.

Why do we need such spaces in QM ? Good question. I believe this mathematical construct has been the most successful enviroment for a theory of atomic (and subatomic) processes.

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Your construction would lead to the L^2 (R^3) which can be shown to be a Hillbert space, actually the single (up to a unitary equivalence) infinite-dimensional separable Hilbert space.
Can you elaborate a bit? In what sense is any complex infinite-dimensional separable Hilbert space equivalent to this one? I looked up "unitary equivalence" and I don't see a definition that applies to Hilbert spaces, but there's one that applies to representations of groups.

Edit: OK, I get it. I forgot for a moment that a Hilbert space isomorphism and a unitary operator is (almost) the same thing. (An isomorphism is a linear surjection $U:X\rightarrow Y$ such that $\langle Ux,Uy\rangle=\langle x,y\rangle$, and a unitary operator is the same thing but with Y=X). So you just meant that all separable infinite-dimensional Hilbert spaces are isomorphic to each other. I actually didn't know that.

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George Jones
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Can you elaborate a bit? In what sense is any complex infinite-dimensional separable Hilbert space equivalent to this one?

Hilbert spaces $H_1$ and $H_2$ are said to be isomorphic if there is an onto linear mapping $U : H_1 \rightarrow H_2$ such $\left< U \psi , U \phi \right>_2 = \left< \psi , \phi \right>_1$ for every $\psi$ and $\phi$ in $H_1$. Such a $U$ is said to be unitary.

Edit Fredrik snuck in with an edit while I was composing the above.

It's not quite true that all Hilbert spaces are isomorphic (unitarily equivalent) to the Hilbert space that you constructed in your first post, but it is true that all separable Hilbert spaces are isomorphic to it. A topological space is said to be separable if it has a countable dense subset. A Hilbert space is separable iff it has a countable orthonormal basis.

From Reed and Simon I:
The following question often puzzles students of functional analysis. If all infinite-dimensional separable Hilbert spaces are the same, why do we talk about them? ... The answer is that we are often interested not just in the space but in some other structures, for example some bounded operators on the space.

Thus, the questions in your original post aren't really affected by this isomorphism property.

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